T=5t^2+40t-35

Solution for T=5t^2+40t-35 equation:

=5T^2+40T-35

We move all terms to the left:

-(5T^2+40T-35)=0

We get rid of parentheses

-5T^2-40T+35=0

a = -5; b = -40; c = +35;
Δ = b2-4ac
Δ = -402-4·(-5)·35
Δ = 2300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2300}=\sqrt{100*23}=\sqrt{100}*\sqrt{23}=10\sqrt{23}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-10\sqrt{23}}{2*-5}=\frac{40-10\sqrt{23}}{-10}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+10\sqrt{23}}{2*-5}=\frac{40+10\sqrt{23}}{-10}
$